Answer:
3 element i.e carbon (C), hydrogen (H) and oxygen.
Explanation:
The following data were obtained from the question:
Subtance >>>>>>>> Chemical Formula
Glucose >>>>>>>>> C₆H₁₂O₆
Methane >>>>>>>> CH₄
Ethanol >>>>>>>>> C₂H₅OH
Hydrogen peroxide >> H₂O₂
From the above table, we can see that ethanol (C₂H₅OH) contains carbon (C), hydrogen (H) and oxygen
Therefore, the total number of elements present in ethanol, C₂H₅OH is 3.
The position of equilibrium lies far to the right, with products being favoured. Hence, option A is correct.
<h3>What is equilibrium?</h3>
Chemical equilibrium is a condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.
A very high value of K indicates that at equilibrium most of the reactants are converted into products.
The equilibrium constant K is the ratio of the concentrations of products to the concentrations of reactants raised to appropriate stoichiometric coefficients.
When the value of the equilibrium constant is very high, the concentration of products is much higher than the concentration of reactants.
This means that most of the reactants are converted into products and the position of equilibrium lies far to the right, with products being favoured.
Hence, option A is correct.
Learn more about the equilibrium here:
brainly.com/question/23641529
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Answer:
C
Explanation:
Characteristics used to classify stars include color, temperature, size, composition, and brightness.
<u>Answer:</u> The standard enthalpy change of the reaction is coming out to be -16.3 kJ
<u>Explanation:</u>
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as 
The equation used to calculate enthalpy change is of a reaction is:
![\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28product%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H_f%28reactant%29%5D)
For the given chemical reaction:
The equation for the enthalpy change of the above reaction is:
![\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCl_2(s))})+(2\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(Mg(OH)_2(s))})+(2\times \Delta H_f_{(HCl(g))})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28MgCl_2%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28H_2O%28g%29%29%7D%29%5D-%5B%281%5Ctimes%20%5CDelta%20H_f_%7B%28Mg%28OH%29_2%28s%29%29%7D%29%2B%282%5Ctimes%20%5CDelta%20H_f_%7B%28HCl%28g%29%29%7D%29%5D)
We are given:
Putting values in above equation, we get:
![\Delta H_{rxn}=[(1\times (-641.8))+(2\times (-241.8))]-[(1\times (-924.5))+(2\times (-92.30))]\\\\\Delta H_{rxn}=-16.3kJ](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B%281%5Ctimes%20%28-641.8%29%29%2B%282%5Ctimes%20%28-241.8%29%29%5D-%5B%281%5Ctimes%20%28-924.5%29%29%2B%282%5Ctimes%20%28-92.30%29%29%5D%5C%5C%5C%5C%5CDelta%20H_%7Brxn%7D%3D-16.3kJ)
Hence, the standard enthalpy change of the reaction is coming out to be -16.3 kJ
